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Question
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.
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Solution
V0 = 81 V
e = 1.6 × 10−19 C
m = 9.1 × 10−31 kg
Maximum kinetic energy of electron,
Kmax = `"eV"_0`
= 1.6 × 10−19 × 81
= 129.6 × 10−19
= 1.29 × 10−17
Kmax = 1.3 × 10−17 J
Maximum velocity of photoelectron,
vmax = `sqrt((2"eV"_0)/"m")`
= `sqrt((2 xx 1.6 xx 10^-19 xx 81)/(9.1 xx 10^-31))`
= `sqrt((259.2 xx 10^-19)/(9.1 xx 10^-31))`
= `sqrt(28.48 xx 10^12)`
vmax = 5.3 × 106 ms−1
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